试题A ：取余

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试题 B: 双阶乘

// 59375 #include <iostream> using namespace std;const int mod = 100000;int main() {int n = 2021;int res = 1;for (int i = 1; i <= 2021; i += 2) {res = (res * i) % mod;}cout << res; }

试题 C: 格点

// 15698 #include <iostream> using namespace std;int main() {int cnt = 0;for (int i = 1; i <= 2021; ++ i) {for (int j = 1; j <= 2021; ++ j) {if (i * j == 2021)cnt ++ ;}}cout << cnt; }

试题 D: 整数分解

• dp
• $f[i][j]$表示用i个数相加表示数字j的方案数

// 691677274345 #include <iostream> #include <cstring> using namespace std;typedef long long ll;ll f[10][2030];int main() {f[0][0] = 1;for (int i = 1; i <= 5; ++ i) {for (int j = 1; j <= 2021; ++ j) {for (int k = 1; k <= 2021; ++ k) {if (j >= k) f[i][j] += f[i - 1][j - k];}}}cout << f[5][2021]; }

• 数论隔板法
• 将2021拆分成2021个1，期间可以放2020个搁板，将2021分成五个正整数相加，也就是在这2020个搁板中选四个位置放四个搁板，因此，答案就是$C_{2020}^4$

#include <iostream> #include <cstring> using namespace std; typedef long long ll;int main() {ll fenmu = 4 * 3 * 2;ll fenzi = 1;for (int i = 2020; i >= 2017; -- i)fenzi *= i;cout << fenzi / fenmu; }

试题 E: 城邦

• prim

// 4046 #include <iostream> #include <cstring> using namespace std; typedef long long ll;int dist[2030]; int g[2030][2030]; bool st[2030];int calc(int x, int y) {int a[5], b[5];for (int i = 0; i < 4; ++ i) {a[i] = x % 10;b[i] = y % 10;x /= 10, y /= 10;}int res = 0;for (int i = 0; i < 4; ++ i) {if (a[i] != b[i])res += a[i] + b[i];}return res; } int prim() {memset(dist, 0x3f, sizeof dist);int res = 0;dist[1] = 0;for (int i = 0; i < 2021; ++ i) {int t = -1;for (int j = 1; j <= 2021; ++ j) {if (!st[j] && (t == -1 || dist[j] < dist[t])) {t = j;}}st[t] = true;res += dist[t];for (int j = 1; j <= 2021; ++ j)dist[j] = min(dist[j], g[t][j]);}return res; }int main() {for (int i = 1; i <= 2021; ++ i) {for (int j = 1; j <= 2021; ++ j) {if (i == j) g[i][j] = g[j][i] = 0x3f3f3f3f;g[i][j] = g[j][i] = calc(i, j);}}cout << prim(); }

• kruscal

#include <iostream> #include <algorithm> using namespace std;const int N = 2021 * 2021 + 10, M = 2021 + 10;struct Edge {int a, b, w;bool operator< (Edge const&e) const {return w < e.w;} }e[N]; int cnt = 0; int p[M];int calc(int x, int y) {int a[5], b[5];for (int i = 0; i < 4; ++ i) {a[i] = x % 10;b[i] = y % 10;x /= 10, y /= 10;}int res = 0;for (int i = 0; i < 4; ++ i) {if (a[i] != b[i])res += a[i] + b[i];}return res; } int find(int x) {if (p[x] != x) p[x] = find(p[x]);return p[x]; } int kruscal() {int res = 0;for (int i = 1; i <= 2021; ++ i)p[i] = i;for (int i = 0; i < cnt; ++ i) {int a = e[i].a, b = e[i].b, w = e[i].w;a = find(a);b = find(b);if (a == b) continue;res += w;p[a] = b;}return res; } int main() {for (int i = 1; i <= 2021; ++ i) {for (int j = 1; j <= 2021; ++ j) {e[cnt ++ ] = {i, j, calc(i, j)};}}sort(e, e + cnt);cout << kruscal(); }

试题 F: 特殊年份

#include <iostream> using namespace std;bool check(int x) {int a[5];for (int i = 0; i < 4; ++ i) {a[i] = x % 10;x /= 10;}if (a[1] == a[3] && a[0] - 1 == a[2]) return true;return false; }int main() {int cnt = 0;for (int i = 0; i < 5; ++ i) {int year;cin >> year;if (check(year))cnt ++ ;}cout << cnt; }

试题 G: 小平方

#include <iostream> using namespace std;int main() {int n;cin >> n;int cnt = 0;for (int i = 1; i <= n - 1; ++ i) {int pf = i * i;if (pf % n < n / 2.0)cnt ++ ;}cout << cnt; }

试题 H: 完全平方数

#include <iostream> #include <cmath> using namespace std;bool check(int x) {int t = (int)sqrt(x);if (t * t == x)return true;return false; }int main() {int n;cin >> n;for (int i = 1; i <= n; ++ i) {if (check(i * n)) {cout << i << endl;return 0;}} }

试题 I: 负载均衡

#include <iostream> #include <queue> using namespace std; typedef pair<int, int> PII; const int N = 2e5 + 10;int power[N]; priority_queue<PII, vector<PII>, greater<PII>> que[N]; // 小顶堆数组int main() {int n, m;cin >> n >> m;for (int i = 1; i <= n; ++ i)cin >> power[i];while (m -- ) {int a, b, c, d;cin >> a >> b >> c >> d;while (que[b].size() && que[b].top().first <= a) {power[b] += que[b].top().second;que[b].pop();}if (power[b] < d) {cout << -1 << endl;} else {power[b] -= d;cout << power[b] << endl;que[b].push({a + c, d});}} }

试题 J: 国际象棋

• 典型的 状压dp
• 由于这里行数比列数少，因此我们枚举每一列的摆放状态
• $f[i][a][b][j]$表示考虑前i列，第i-1列状态为a，第i列状态为b，且一共放了j个马（因为这道题与数量有关）的方案数
• 确定正确的初始状态

#include <iostream> using namespace std; const int N = 110, M = (1 << 6), K = 21, MOD = 1e9 + 7;int f[N][M][M][K];int get_count(int x) { // 二进制下1的个数int res = 0;while (x) {res ++ ;x -= (x & -x);}return res; }int main() {int n, m, k;cin >> n >> m >> k;f[0][0][0][0] = 1; // 初始状态for (int i = 1; i <= m; ++ i) { // 枚举每一列for (int a = 0; a < (1 << n); ++ a) { // 第i-2列for (int b = 0; b < (1 << n); ++ b) { // 第i-1列if ((a & (b << 2)) || ((a << 2) & b)) continue; // 状态矛盾for (int c = 0; c < (1 << n); ++ c) { // 第i列if ((c & (b << 2)) || ((c << 2) & b)) continue; // 状态矛盾if ((c & (a << 1)) || ((c << 1) & a)) continue; // 状态矛盾int t = get_count(c); // 根据第i列个数来确定总个数的下限for (int j = t; j <= k; ++ j) { // 枚举总个数f[i][b][c][j] = (f[i][b][c][j] + f[i - 1][a][b][j - t]) % MOD;}}}}}int res = 0;for (int i = 0; i < (1 << n); ++ i) {for (int j = 0; j < (1 << n); ++ j) {res = (res + f[m][i][j][k]) % MOD;}}cout << res; }

总结

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